A 661-N force is being applied to a 121-kg block. The coefficient of static friction between the block and the surface is 0.410. What is the minimal amount of additional mass that must be added on top of the block to prevent the block from moving?
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A 661-N force is being applied to a 121-kg block. The coefficient of static friction between the block and the surface is 0.410. What is the minimal amount of additional mass that must be added on top of the block to prevent the block from moving?
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
The frictional force is the coefficient of friction times the normal force.
Here, you want the frictional force to equal 661, so there is no movement.
So:\[F_{fr} = \mu_s \times F_N \rightarrow 661 = 0.410 \times mg\]
From this, you can solve for mg, and you get mg = 1612
dividing by g (9.8), you get m = 164.5.
So, since you already have 121 kg, you only need 43.5 more.